$A$ $4.5 \;cm$ needle is placed $12 \;cm$ away from a convex mirror of focal length $15 \;cm$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

(N/A) Height of the needle,$h_{1} = 4.5 \; cm$.
Object distance,$u = -12 \; cm$.
Focal length of the convex mirror,$f = 15 \; cm$.
Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-12}) = \frac{1}{15} + \frac{1}{12} = \frac{4+5}{60} = \frac{9}{60}$.
Thus,$v = \frac{60}{9} \approx 6.7 \; cm$.
The image is formed $6.7 \; cm$ behind the mirror.
Magnification $m = -\frac{v}{u} = -\frac{6.7}{-12} \approx 0.56$.
The height of the image $h_{2} = m \times h_{1} = 0.56 \times 4.5 \approx 2.5 \; cm$.
The image is virtual,erect,and diminished.
As the needle is moved farther from the mirror,the object distance $u$ increases,causing the image distance $v$ to increase towards the focus $f$,and the size of the image decreases.